Evaluate: `sqrt(-2+2sqrt(3)i)`

Options

A. `pm (1+ sqrt(3)i)`
B. `pm (1- sqrt(3)i)`
C. `pm (-1+ sqrt(3)i)`
D. None of these

Answer

To evaluate \( \sqrt{-2 + 2\sqrt{3}i} \), we will assume that it can be expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers.

### Step 1: Set up the equation
Assume:
\[ \sqrt{-2 + 2\sqrt{3}i} = a + bi \]

### Step 2: Square both sides
Squaring both sides gives:
\[ -2 + 2\sqrt{3}i = (a + bi)^2 \]
Expanding the right side:
\[ (a + bi)^2 = a^2 + 2abi - b^2 = (a^2 - b^2) + (2ab)i \]
Thus, we have:
\[ -2 + 2\sqrt{3}i = (a^2 - b^2) + (2ab)i \]

### Step 3: Equate real and imaginary parts
From the equation, we can equate the real and imaginary parts:
1. \( a^2 - b^2 = -2 \) (Real part)
2. \( 2ab = 2\sqrt{3} \) (Imaginary part)

### Step 4: Simplify the equations
From the second equation:
\[ ab = \sqrt{3} \]
Now we can express \( a \) in terms of \( b \):
\[ a = \frac{\sqrt{3}}{b} \]

### Step 5: Substitute into the first equation
Substituting \( a \) into the first equation:
\[ \left(\frac{\sqrt{3}}{b}\right)^2 - b^2 = -2 \]
This simplifies to:
\[ \frac{3}{b^2} - b^2 = -2 \]
Multiplying through by \( b^2 \) to eliminate the fraction:
\[ 3 - b^4 = -2b^2 \]
Rearranging gives:
\[ b^4 - 2b^2 - 3 = 0 \]

### Step 6: Let \( x = b^2 \)
Let \( x = b^2 \), then we have a quadratic equation:
\[ x^2 - 2x - 3 = 0 \]

### Step 7: Factor the quadratic
Factoring gives:
\[ (x - 3)(x + 1) = 0 \]
Thus, \( x = 3 \) or \( x = -1 \). Since \( x = b^2 \) must be non-negative, we have:
\[ b^2 = 3 \implies b = \pm \sqrt{3} \]

### Step 8: Find \( a \)
Using \( b = \sqrt{3} \):
\[ a = \frac{\sqrt{3}}{\sqrt{3}} = 1 \]
Using \( b = -\sqrt{3} \):
\[ a = \frac{\sqrt{3}}{-\sqrt{3}} = -1 \]

### Step 9: Write the final result
Thus, we have two possible values for \( \sqrt{-2 + 2\sqrt{3}i} \):
\[ \sqrt{-2 + 2\sqrt{3}i} = 1 + \sqrt{3}i \quad \text{or} \quad -1 - \sqrt{3}i \]

### Final Answer
The final answer is:
\[ \sqrt{-2 + 2\sqrt{3}i} = \pm (1 + \sqrt{3}i) \]